DP on Trees Diameter of Tree in DSA Rust - Time & Space Complexity

Time Complexity: DP on Trees Diameter of Tree

O(n)

Understanding Time Complexity

We want to understand how long it takes to find the longest path in a tree using dynamic programming.

How does the time needed grow as the tree gets bigger?

Scenario Under Consideration

Analyze the time complexity of the following code snippet.

fn dfs(node: usize, parent: usize, graph: &Vec<Vec<usize>>, max_diameter: &mut usize) -> usize {
    let mut max1 = 0;
    let mut max2 = 0;
    for &child in &graph[node] {
        if child == parent { continue; }
        let depth = dfs(child, node, graph, max_diameter) + 1;
        if depth > max1 {
            max2 = max1;
            max1 = depth;
        } else if depth > max2 {
            max2 = depth;
        }
    }
    *max_diameter = (*max_diameter).max(max1 + max2);
    max1
}
    

This code finds the diameter of a tree by exploring each node once and tracking the longest paths.

Identify Repeating Operations

Identify the loops, recursion, array traversals that repeat.

• Primary operation: Recursive depth-first search (DFS) visiting each node.
• How many times: Each node is visited exactly once, and each edge is checked once.

How Execution Grows With Input

As the number of nodes grows, the number of operations grows roughly the same amount.

Input Size (n)	Approx. Operations
10	About 10 visits and 9 edge checks
100	About 100 visits and 99 edge checks
1000	About 1000 visits and 999 edge checks

Pattern observation: The work grows linearly with the number of nodes.

Final Time Complexity

Time Complexity: O(n)

This means the time to find the diameter grows directly in proportion to the number of nodes in the tree.

Common Mistake

[X] Wrong: "Since the code uses recursion inside a loop, it must be quadratic time O(n²)."

[OK] Correct: Each node is visited only once because the recursion skips the parent, so the total work is linear, not quadratic.

Interview Connect

Understanding this linear time approach to tree diameter shows you can handle tree problems efficiently, a key skill in many coding challenges.

Self-Check

"What if we changed the tree to a graph with cycles? How would the time complexity change?"

Practice

1. What does the diameter of a tree represent in the context of DP on Trees?

easy

A. The longest path between any two nodes in the tree

B. The number of nodes in the tree

C. The height of the tree from the root

D. The sum of all edge weights in the tree

Solution

1. Step 1: Understand the definition of tree diameter

   The diameter is the longest path between any two nodes, not just from the root or sum of edges.

2. Step 2: Differentiate from other tree properties

   Height is from root to leaf, sum of edges is total weight, number of nodes is size, none equal diameter.

3. Final Answer:

   The longest path between any two nodes in the tree -> Option A

4. Quick Check:

   Diameter = Longest path [OK]

Hint: Diameter = longest path between any two nodes [OK]

Common Mistakes:

• Confusing diameter with tree height
• Thinking diameter is total nodes count
• Assuming diameter is sum of all edges

2. Which Rust syntax correctly declares a recursive function to compute the diameter of a tree represented as adjacency list Vec?

easy

A. fn diameter(node: usize, parent: usize, graph: &mut Vec) -> usize { ... }

B. fn diameter(node: usize, graph: Vec) -> usize { ... }

C. fn diameter(node: usize, parent: usize, graph: &Vec) -> usize { ... }

D. fn diameter(&self, node: usize, parent: usize) -> usize { ... }

Solution

1. Step 1: Identify correct parameter passing for graph

   Graph should be passed by reference (&Vec) to avoid copying large data.

2. Step 2: Include parent parameter to avoid revisiting parent node

   Parent node is needed to prevent cycles in undirected tree traversal.

3. Final Answer:

   fn diameter(node: usize, parent: usize, graph: &Vec) -> usize { ... } -> Option C

4. Quick Check:

   Pass graph by reference and track parent [OK]

Hint: Pass graph as &Vec and track parent node [OK]

Common Mistakes:

• Passing graph by value causing performance issues
• Missing parent parameter causing infinite recursion
• Using &mut when mutation is not needed

3. Given the following Rust code snippet to compute the diameter of a tree, what is the printed output?

fn dfs(node: usize, parent: usize, graph: &Vec, max_diameter: &mut usize) -> usize {
    let mut max1 = 0;
    let mut max2 = 0;
    for &child in &graph[node] {
        if child == parent { continue; }
        let height = dfs(child, node, graph, max_diameter);
        if height > max1 {
            max2 = max1;
            max1 = height;
        } else if height > max2 {
            max2 = height;
        }
    }
    *max_diameter = (*max_diameter).max(max1 + max2);
    max1 + 1
}

fn main() {
    let graph = vec![
        vec![1, 2],
        vec![0, 3, 4],
        vec![0],
        vec![1],
        vec![1]
    ];
    let mut max_diameter = 0;
    dfs(0, usize::MAX, &graph, &mut max_diameter);
    println!("{}", max_diameter);
}

medium

A. 3

B. 4

C. 2

D. 5

Solution

1. Step 1: Trace DFS heights from leaves

   Leaves 2, 3, 4 return height 1; node 1 gets max1=1, max2=1; node 0 gets max1=2 (from node 1), max2=1 (from node 2).

2. Step 2: Calculate max diameter updates

   At node 1, max_diameter = max(0,1+1)=2; at node 0, max_diameter = max(2,2+1)=3.

3. Final Answer:

   3 -> Option A

4. Quick Check:

   Longest path length = 3 edges [OK]

Hint: Sum two largest subtree heights at each node [OK]

Common Mistakes:

• Counting nodes instead of edges for diameter
• Not updating max_diameter correctly
• Ignoring parent to avoid revisiting

4. Identify the bug in this Rust function computing tree diameter and select the fix:

fn dfs(node: usize, graph: &Vec, max_diameter: &mut usize) -> usize {
    let mut max1 = 0;
    let mut max2 = 0;
    for &child in &graph[node] {
        let height = dfs(child, graph, max_diameter);
        if height > max1 {
            max2 = max1;
            max1 = height;
        } else if height > max2 {
            max2 = height;
        }
    }
    *max_diameter = (*max_diameter).max(max1 + max2);
    max1 + 1
}

medium

A. Initialize max1 and max2 to 1 instead of 0

B. Add a parent parameter to avoid revisiting the parent node

C. Change max_diameter to return instead of using mutable reference

D. Remove max2 variable as it is unnecessary

Solution

1. Step 1: Identify infinite recursion cause

   Without parent tracking, dfs revisits parent causing infinite recursion or stack overflow.

2. Step 2: Fix by adding parent parameter

   Pass parent node to skip revisiting it in children loop, preventing cycles.

3. Final Answer:

   Add a parent parameter to avoid revisiting the parent node -> Option B

4. Quick Check:

   Parent tracking prevents infinite recursion [OK]

Hint: Always track parent node in undirected tree DFS [OK]

Common Mistakes:

• Not passing parent causing infinite recursion
• Misusing mutable references for max_diameter
• Incorrect initialization of max heights

5. You want to find the diameter of a weighted tree where edges have weights. Which modification to the DP on Trees diameter algorithm is correct?

hard

A. Ignore weights and compute diameter as number of edges only

B. Sum all edge weights and divide by two to get diameter

C. Use BFS twice without any DP or recursion

D. Store and propagate the maximum weighted path length instead of just height counts

Solution

1. Step 1: Understand weighted tree diameter difference

   Diameter must consider edge weights, so heights become max weighted path lengths.

2. Step 2: Modify DP to accumulate weights

   Instead of counting edges, add edge weights when computing subtree heights and update max diameter accordingly.

3. Final Answer:

   Store and propagate the maximum weighted path length instead of just height counts -> Option D

4. Quick Check:

   Weighted diameter needs weighted path sums [OK]

Hint: Accumulate edge weights in DP for weighted diameter [OK]

Common Mistakes:

• Ignoring weights leads to wrong diameter
• Using BFS without weights misses weighted paths
• Incorrect formula dividing total weights